The stopping potentials are V1 and V2 with incident lights of wavelength $λ_1$ and $λ_2$ respectively. Then V1 – V2
Answer & explanation
Correct answer: option 2
$eV_1=\frac{hc}{λ_1}-\phi$
$eV_2=\frac{hc}{λ_2}-\phi$
$V_1-V_2=\frac{hc}{e}(\frac{1}{λ_1}-\frac{1}{λ_2})$